Question 136498
Hi. I've been trying to solve this for about three hours. I know the solution is 11.0834645967, but I can't figure out how to get it within the constraints of the problem: Show how to solve the following equation in terns of the common logarithm. Also give an approximation of the solution to the nearest hundreth of a unit. 

{{{3^(2x-3)=5^(x+2)}}} 

My work is 

{{{  log((  3^(2x-3) )) = log((  5^(x+2)  ))  }}} 

   {{{ (2x-3)log((3)) = (x+2)log((5))}}} 

          {{{2x-3=(x+2)(log((5))/log((3)))}}}
<pre><b>
You did fine to here.  Then you went astray. Now here is what 
you should have done:

Get your calculator at this point and calculate {{{(log((5))/log((3)))}}} as

{{{1.464973521}}}.

and store it in the memory of your calculator, or write it down.
So you won't have to cary such a long awkward decimal through your
steps, just call it {{{A}}}.  In other words let {{{A = (log((5))/log((3)))}}}
{{{1.464973521}}}

Now 

          {{{2x-3=(x+2)(log((5))/log((3)))}}}

becomes simply

          {{{2x-3=(x+2)A}}}

or switch the A to the left of the parentheses:

          {{{2x-3=A(x+2)}}}

Distribute on the right:

          {{{2x-3=Ax+2A}}}

Get all the x terms on the left and other terms on the right:

          {{{2x-Ax=3+2A}}}

Factor out x on the left:

          {{{x(2-A)=3+2A}}}

Divide both sides by {{{(2-A)}}}

          {{{(x(2-A))/((2-A))=(3+2A)/((2-A))}}}

Cancel the {{{2-A}}}'s

          {{{x*cross((2-A))/cross((2-A))=(3+2A)/((2-A))}}}

          {{{x=(3+2A)/(2-A)}}}

Now using the recall feature of your calculator to substitute
the value of A.

          {{{x=(3+2(1.464973521))/(2-1.464973521)}}}

Now calculate that with your calculator, and you'll get

          {{{x=11.0834645967}}}

Edwin</pre>